Question

For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 bar for all species. For the reaction
2 NO ( g ) + O 2 ( g ) -------> 2 NO 2 ( g )
the standard change in Gibbs free energy is Δ G ° = − 32.8 kJ / mol . What is Δ G for this reaction at 298 K when the partial pressures are:
PNO = 0.500 bar , PO2 = 0.250 bar , and PNO 2 = 0.800 bar
DeltaG = ?

Answer 1

Answer : The value of $\Delta G_{rxn}$ is, -27.0kJ/mole

Explanation :

The formula used for $\Delta G_{rxn}$ is:

$\Delta G_{rxn}=\Delta G^o+RT\ln K_p$   ............(1)

where,

$\Delta G_{rxn}$ = Gibbs free energy for the reaction

$\Delta G_^o$ =  standard Gibbs free energy  = -32.8 kJ

/mol

R = gas constant = 8.314 J/mole.K

T = temperature = 298 K

$K_p$ = equilibrium constant

First we have to calculate the value of $K_p$.

The given balanced chemical reaction is,

$2NO(g)+O_2(g)\rightarrow 2NO_2(g)$

The expression for equilibrium constant will be :

$K_p=\frac{(p_{NO_2})^2}{(p_{NO})^2\times (p_{O_2})}$

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

$K_p=\frac{(0.800)^2}{(0.500)^2\times (0.250)}$

$K_p=10.24$

Now we have to calculate the value of $\Delta G_{rxn}$ by using relation (1).

$\Delta G_{rxn}=\Delta G^o+RT\ln K_p$

Now put all the given values in this formula, we get:

$\Delta G_{rxn}=-32.8kJ/mol+(8.314\times 10^{-3}kJ/mole.K)\times (298K)\ln (10.24)$

$\Delta G_{rxn}=-27.0kJ/mol$

Therefore, the value of $\Delta G_{rxn}$ is, -27.0kJ/mole